find a formula for f'(x) and determine the slope f'(a) at the point where x=a is given.

f(x)= 5*e^x/2 ; x=0

Question

find a formula for f'(x) and determine the slope f'(a) at the point where x=a is given.


f(x)= 5*e^x/2  ; x=0

e.mccormick · Answer

"slope f'(a) at" so the slope of the derivative?  Or is that supposed to be "slope f(a) at"?

e.mccormick · Answer

Ah, ok...  I added an OF there mentally.

e.mccormick · Answer

OK, $f'(x)$ will give you the slope, so what you really need is $f'(x)$.  It is the answer to the first part and finds the second.

e.mccormick · Answer

So, what is the derivative of $\Large5e^{\frac{x}{2}}$?  Do you know how to find that one.

anonymous · Answer

I have no idea because of the e^x/2. what formula would I use?

e.mccormick · Answer

OK, so, do you know the derivative of just $e^x$?

anonymous · Answer

no

e.mccormick · Answer

OK.  That one is easy.  The derivative of $e^x$ is $e^x$.  It is special.  It is its own derivative.

anonymous · Answer

so how would I find e^x/2

e.mccormick · Answer

This will be a chain rule.  You have $f(x)^{g(x)}$.  Now the $f(x)$ part is $5e^{x/2}$ and the $g(x)$ part is $\frac{x}{2}$.

e.mccormick · Answer

The chain rule is for compositions of functions.  Because $e^x$ is a function and the power is made of another function, you do the chain rule.  Do you remember the chain rule?

anonymous · Answer

is it e^(g(x)) * g '(x)  ?

e.mccormick · Answer

!!  Exactly!

anonymous · Answer

but how do I find g '(x) if g(x)= x/2?

e.mccormick · Answer

Well, it is a variable over a constant, but there is another way to think about that.  When you multiply x by a fraction, what happens to it?  And can you undo the multiplication?

anonymous · Answer

im confused...

e.mccormick · Answer

$$5\cdot \frac{1}{3}=\frac{5}{3}$$So what if that 5 was your x and that 3 was your 2?

anonymous · Answer

x*1/2?

e.mccormick · Answer

Yep!  It just hopps off the /2 and makes a half times x.

anonymous · Answer

so would the equation be 5e^x/2 * 1/2x ?

e.mccormick · Answer

Almost, you need to differentiate the $\frac{1}{2}x$

anonymous · Answer

so 1/2

e.mccormick · Answer

/cheer Yep.

e.mccormick · Answer

And since the differentiatoion of the e part is itself, it is just the original equation, but now over 2!

anonymous · Answer

so f '(x)=(5e^x/2) / 2

e.mccormick · Answer

Yes.  That is the first half of this.

When it comes to differentiation and integration, e in $e^x$ stands for easy!  It is itself.  If there is anything other than just an x in the exponent, you chain rule it.  And if you somehow get $e^c$ where c is any constant, that is just a number so the rules for constants apply.

e.mccormick · Answer

So now they want you to put 0 in for x.

anonymous · Answer

would that cancel out the e since its zero?

e.mccormick · Answer

That is a good way to say it.  It makes value 1.

e.mccormick · Answer

$$\frac{5e^{\frac{x}{2}}}{2}\implies \frac{5e^{\frac{0}{2}}}{2}\implies \frac{5e^0}{2}\implies \frac{5\cdot 1}{2}\implies \frac{5}{2}$$

e.mccormick · Answer

Which is exactly what you said... just all written out in math stuff.  Hehe.

anonymous · Answer

ok I think I have it now thanx for your patience

e.mccormick · Answer

No problem.  You just needed a little help with e and a reminder on the meaning of a fraction.  Work with some more e ones and you will remember that part.  The fraction... well... calculus uses every piece of math before it.  Want to do good in calculus?  Practice everything that came before it!

e.mccormick · Answer

I am taking an intro to Linear Algebra class at the moment... final on Monday... and it is tons and tons of basic algebra all wrapped around a few new concepts.  Where everyone has problems is with dropping a sign, doing the algebra wrong, and so on.